Find the eccentricity, co ordinates of foci-length of latus rectum and equation of directrices of the folloeing ellipses.
`9x^2+16y^2-36x+32y-92=0`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the given equation
The given equation is:
\[ 9x^2 + 16y^2 - 36x + 32y - 92 = 0 \]

### Step 2: Rearrange the equation
We will rearrange the equation to group the \(x\) and \(y\) terms:
\[ 9x^2 - 36x + 16y^2 + 32y = 92 \]

### Step 3: Complete the square for \(x\) and \(y\)
For the \(x\) terms:
\[ 9(x^2 - 4x) \]
To complete the square, we take half of \(-4\) (which is \(-2\)), square it (getting \(4\)), and add and subtract it inside the parentheses:
\[ 9((x - 2)^2 - 4) = 9(x - 2)^2 - 36 \]

For the \(y\) terms:
\[ 16(y^2 + 2y) \]
Again, we take half of \(2\) (which is \(1\)), square it (getting \(1\)), and add and subtract it:
\[ 16((y + 1)^2 - 1) = 16(y + 1)^2 - 16 \]

Now substituting back into the equation:
\[ 9(x - 2)^2 - 36 + 16(y + 1)^2 - 16 = 92 \]

### Step 4: Simplify the equation
Combine the constants:
\[ 9(x - 2)^2 + 16(y + 1)^2 - 52 = 92 \]
\[ 9(x - 2)^2 + 16(y + 1)^2 = 144 \]

### Step 5: Divide by 144 to get the standard form
\[ \frac{(x - 2)^2}{16} + \frac{(y + 1)^2}{9} = 1 \]

### Step 6: Identify \(a^2\) and \(b^2\)
From the standard form, we identify:
- \(a^2 = 16\) (thus \(a = 4\))
- \(b^2 = 9\) (thus \(b = 3\))

### Step 7: Find the eccentricity \(e\)
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 8: Find the coordinates of the foci
The foci are located at \((h \pm ae, k)\):
- Here, \(h = 2\), \(k = -1\), and \(ae = 4 \cdot \frac{\sqrt{7}}{4} = \sqrt{7}\).
Thus, the coordinates of the foci are:
\[ (2 \pm \sqrt{7}, -1) \]

### Step 9: Find the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} \]
Substituting the values:
\[ L = \frac{2 \cdot 9}{4} = \frac{18}{4} = 4.5 \]

### Step 10: Find the equations of the directrices
The equations of the directrices are given by:
\[ x = h \pm \frac{a}{e} \]
Calculating \(\frac{a}{e}\):
\[ \frac{a}{e} = \frac{4}{\frac{\sqrt{7}}{4}} = \frac{16}{\sqrt{7}} \]
Thus, the equations of the directrices are:
\[ x = 2 \pm \frac{16}{\sqrt{7}} \]

### Summary of Results
- Eccentricity \(e = \frac{\sqrt{7}}{4}\)
- Coordinates of foci: \((2 + \sqrt{7}, -1)\) and \((2 - \sqrt{7}, -1)\)
- Length of the latus rectum: \(4.5\)
- Equations of the directrices: \(x = 2 + \frac{16}{\sqrt{7}}\) and \(x = 2 - \frac{16}{\sqrt{7}}\)